2
$\begingroup$

Let $n \in \mathbb{N}$, $K$ a finite field. Denote by $K[[x]]$ the (profinite) ring of formal power series over $K$. Note that $\text{GL}_n(K[[x]])$ is a profinite group.

Is every closed subgroup of $\text{GL}_n(K[[x]])$ (topologically) finitely generated?

An analogy is that for evey prime number $p$, every closed subgroup of $\text{GL}_n(\mathbb{Z}_p)$ is finitely generated (in fact there is a uniform bound on the number of generators for all of the closed subgroups, and I expect to have such a bound here as well).

$\endgroup$
5
  • $\begingroup$ Note that the question is about $K[[x]]$ and not $K((x))$. $\endgroup$
    – Pablo
    Commented Aug 9, 2014 at 17:35
  • 3
    $\begingroup$ This is false for $n=1$ and the entire group. It suffices to show that for the local field $F = K(\!(x)\!)$, $F^{\times}$ has infinitely many non-trivial continuous homomorphisms to $\mathbf{Z}/p\mathbf{Z}$. By local class field theory, it is the same to say that ${\rm{H}}^1(F, \mathbf{Z}/p\mathbf{Z})$ is infinite. By the Artin-Schreier sequence, this cohomology is the same as the continuous $\mathbf{F}_p$-dual of the pro-$p$ group $F/\wp(F)$ where $\wp(f)=f^p-f$. By considering polar parts, $F/\wp(F)$ is infinite. $\endgroup$
    – user27920
    Commented Aug 9, 2014 at 17:36
  • $\begingroup$ Since $K(\!(x)\!)^{\times} = x^{\mathbf{Z}} \times K[\![x]\!]^{\times}$ as topological groups, to show the answer is negative for $n=1$ and the whole group it suffices to show that $K(\!(x)\!)^{\times}$ has infinitely many continuous homomorphisms to $\mathbf{Z}/p\mathbf{Z}$, as is done in my preceding comment. And then via determinant it fails (on ${\rm{GL}}_n(K[\![x]\!])$) for all $n$. The question should perhaps be posed for SL$_n$. $\endgroup$
    – user27920
    Commented Aug 9, 2014 at 17:44
  • 1
    $\begingroup$ I would gladly accept this as an answer. Could you please elaborate more on the connection between $K((x))$ and $K[[x]]$ and how the solution for the former gives a solution for the latter? $\endgroup$
    – Pablo
    Commented Aug 9, 2014 at 18:21
  • 1
    $\begingroup$ An explicit indication of infinitely many polar parts that are not equal in $F/\wp(F)$ is mentioned in math.stackexchange.com/questions/353928/…. $\endgroup$
    – KConrad
    Commented Aug 10, 2014 at 6:39

1 Answer 1

13
$\begingroup$

This question has a negative answer is many respects. Firstly, there are simple constructions in the commutative case. Namely, the additive group $K[\![x]\!]$ is an infinite dimensional $\mathbb F_p$-vector space and thus not finitely generated. Less simply, the multiplicative group $K[\![x]\!]^\times$ has infinitely many homomorphisms to $\mathbb F_p$. That is, the quotient by the group of $p$-th powers is infinite dimensional. The group of $p$-th powers is easy to calculate because the $p$-th power is Frobenius, a ring homomorphism: it is the power series of the form $\sum a_n t^{np}$, $a_0\ne0$. Thus the elements $1+t^n$, for $n$ not divisible by $p$, forms a linearly independent sequence in the $\mathbb F_p$-vector space of the mod $p$ quotient. Class field theory relates the slightly bigger group of the mod $p$ quotient of $K((x))^\times$ with the abelianized Galois group of $K((x))$ and with the quotient of the additive group by Artin-Schreier map $f\mapsto f^p-f$.

Thus for any $n$ the whole group $\rm{GL}_n(K[\![x]\!])$ is not finitely generated because it surjects via the determinant to the multiplicative group, which is not. So one might restrict to $\rm{SL}_n$. But this contains subgroups that are not semisimple, such as $\mathbb G_a$, $\mathbb G_m$, $\rm{GL}_{n-1}$, and the Borel subgroup. None of these are finitely generated, but all for the same commutative reason. But this is really cheating: clearly the "correct" question should get away from the silly constructions that one can make with ease in the commutative setting, so one ought to replace GL$_n$ with semisimple groups $G$ over $O = K[\![x]\!]$.

Likewise, if $G$ is not simply connected then more commutative silliness gets in the way. For example, the determinant induces ${\rm{PGL}}_n(O) \twoheadrightarrow O^{\times}/(O^{\times})^n$, so for $n$ divisible by $p$ we have the same problem once again (as $O^{\times}/(O^{\times})^p$ is a quotient, and it has infinitely many continuous homomorphisms to $\mathbf{Z}/p\mathbf{Z}$ as we have seen above).

So finally it seems that the version of the question not easily falsified by commutative tricks is for $G$ to be a semisimple $O$-group that is simply connected and to consider the open subgroups of $G(O)$ (which are of course automatically closed). It is a general fact (via deformation theory of semisimple groups) that any such $G$ is the scalar extension along $K \rightarrow O$ of its reduction, though you might wish to just consider only such $G$ without knowing it is the most general case. By coming from $K$ we see that $G$ arises from a "constant" $K$-group over the global ring $K[x]$. One can then use strong approximation for simply connected groups over global fields (such as $K(x)$) and results of Behr on finite generation of $S$-arithmetic groups when the sum of local ranks at $v \in S$ is at least 3) to deduce that $G(O)$ contains a dense finitely generated subgroup, namely an $S$-arithmetic group for suitable $S$, so $G(O)$ is topologically finitely generated. Moreover, any open subgroup of $G(O)$ is defined by a "congruence condition", so one can still find $S$-arithmetic groups dense in such open subgroups, and hence they're again topologically finitely generated.

$\endgroup$
1
  • 1
    $\begingroup$ Relevant to the last paragraph: an open subgroup of a finitely generated profinite group is finitely generated. This is closely related to: a finite index subgroup of a finitely generated group is finitely generated. $\endgroup$ Commented Aug 10, 2014 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.